[Sally Le Page (OP)]

David asked us:

"I have been trying to work out the difference between the Earth's polar radius and its equatorial radius, just based on a very simple model (hydrostatic equilibrium with gravity and centrifugal effects only). Apparently Newton was able to do this in the Principia, and got close to the actual value for the Earth. Problem is when I do it I always get about half the correct difference (about 10km as opposed to about 21km for the actual value). So my question is, are there factors determining the shape of the Earth (geological forces, etc.) that would make that much difference, or am I just doing something else wrong in my calculations? Or to put it another way: if I do this based on a simple hydrostatic model should I be getting something better than half the true value?"

Can anyone help?

[Bored chemist]

In theory, you can simply look up how Newton did it.
https://www.gutenberg.org/ebooks/28233

[evan_au]

In practise, I can't read Latin...

[Colin2B]

David asked us:
“Apparently Newton was able to do this in the Principia, and got close to the actual value for the Earth. ............ Problem is when I do it I always get about half the correct difference (about 10km as opposed to about 21km for the actual value). “

I’m not sure he worked it out in detail in the Principia, I’ve not seen a formula there. He described the method and worked out the flattening ratio where e=equatorial radius, p=polar radius, but using assumptions from the time he was out, compared to modern measurements, by about 30%. A discrepancy of 2x suggests you might have used diameter rather than radius, but if you calculated via the flattening ratio that shouldn’t be a problem.
What ratio do you get? I think he got 1/230. Do you want to show your working?

[Bored chemist]

In practise, I can't read Latin...

It may startle you to discover this, but there are translations available.

[dvilla]

I’m not sure he worked it out in detail in the Principia, I’ve not seen a formula there. He described the method and worked out the flattening ratio where e=equatorial radius, p=polar radius, but using assumptions from the time he was out, compared to modern measurements, by about 30%. A discrepancy of 2x suggests you might have used diameter rather than radius, but if you calculated via the flattening ratio that shouldn’t be a problem.
What ratio do you get? I think he got 1/230. Do you want to show your working?

I get a ratio of about 1/582.
I refer to the following website:-

    www . sjsu . edu / faculty / watkins / earthshape.htm  (sorry about the spaces - this editor automatically creates links and does not let me post them!)

Essentially, I concur with everything the authors of this site say (equations, values, methodology, etc) EXCEPT when they say "The error in terms of the difference in the RHS versus the LHS is small, about 0.3 of 1 percent" they are implying that the error is small. But what they are comparing is the difference in relation to the overall radius of Earth. What they should be comparing is the the difference between the calculated value and the actual value of the "bulge" in relation to the actual size of the "bulge". This error is more like 50%.
  The question is - is this difference more than you would expect once other factors are taken into account? I think it is, as I can't think of what other factors would be in play that would account for it. Sure there are bound to be geological forces and tidal forces, and the gravitation field of Earth itself is not perfectly spherically symmetric, but are these enough to make this much difference? (eg, tidal forces would account for a difference of only a few metres)

[Colin2B]

@dvilla
Welcome to the forum, thanks for registering. I’ll take a look at that reference and get back to you, but 582 is definitely too big.

[Colin2B]

I get a ratio of about 1/582.
I refer to the following website:-

I’ve had a look at that and the author is not using Newton’s method. Newton described two columns of water, one on the polar axis and one one the equatorial, connected at the earth centre. He considers that the weight of the 2 columns must be equal and their height a result of gravity for the polar direction and the equilibrium of gravitational and centrifugal forces for equatorial direction. In the Principia he says he made some assumptions based on measurements of the ratio of gravitational to centrifugal taken at Paris.
The link’s author is looking at the equipotential surface for a rotating fluid sphere and is making the assumption that the earth has uniform density. The real earth has a very dense core and a surface which can behave like a fluid - even the moon causes a solid tide! We also know that the geoid - the equivalent of sea level - varies significantly due to the effect of subsurface density changes. All of this means that a simple model will always be out compared to the real earth, but this particular model is even further out than Newton’s.

Chandrasekhar, who studied at Cambridge, made a translation of the Principia which you might like to follow up. I’m also sure he did some work on the earth shape which might be worth finding.

[Just thinking]

I think that is impossible to ever know with great accuracy as the elasticity of the earth can not be accounted for but I'm sure that the equator is a greater distance than around the poles.

[Bored chemist]

I think that is impossible to ever know with great accuracy

We can measure it.
Also, given how long it has had to settle and the fact that most of it is covered with water, we can assume it's a liquid with no elasticity.

[Just thinking]

We can measure it.
Also, given how long it has had to settle and the fact that most of it is covered with water, we can assume it's a liquid with no elasticity.

I guess that South Africa is the largest landmass on the equator and as you say the rest is water witch will find its own level. Having said that the ocean will be spun out at the equator but not to the extent of the land. And of course, the moon and the sun will play a part in the equation. So the overall increase in the earth diameter around the equator is for the most part water.

[Colin2B]

Having said that the ocean will be spun out at the equator but not to the extent of the land.......So the overall increase in the earth diameter around the equator is for the most part water.

Not so. As @bored said there has been a long time for things to settle and the average land surface will also flow to an equilibrium position along with the sea, so the land at the equator is therefore higher that it would have been at the equator by the same amount as the sea would be.

And of course, the moon and the sun will play a part in the equation.

Again no. The sun and moon do not act directly over the equator and the effects are transitory so do not affect the average level.
Satellite and aerial surveys are giving very accurate measurements. The Grav D project will allow surface height measurements to within 2cm.
It’s worth noting that the effective sea level (called the geoid) varies over the surface of the earth due to local variations in density.

[Just thinking]

Quote from: Just thinking on Yesterday at 08:11:20

    Having said that the ocean will be spun out at the equator but not to the extent of the land.......So the overall increase in the earth diameter around the equator is for the most part water.

Not so. As @bored said there has been a long time for things to settle and the average land surface will also flow to an equilibrium position along with the sea, so the land at the equator is therefore higher that it would have been at the equator by the same amount as the sea would be.

Quote from: Just thinking on Yesterday at 08:11:20

    And of course, the moon and the sun will play a part in the equation.

Again no. The sun and moon do not act directly over the equator and the effects are transitory so do not affect the average level.
Satellite and aerial surveys are giving very accurate measurements. The Grav D project will allow surface height measurements to within 2cm.
It’s worth noting that the effective sea level (called the geoid) varies over the surface of the earth due to local variations in density.

It's calculated that from east to west there is a 170-kilometre increase in the distance what that equates to in altitude I'm not sure. Your explanation makes good sense apparently they can measure the distance to the moon with great accuracy using a laser beam and mirrors on the moon. this Grav D project you mentioned sounds very interesting as there would be many varying factors to consider like decaying orbit the moon and the sun the overall altitude of the satellite would be the big challenge.